Question:

Let u = sec x + tan x. Then du = sec x (sec x + tan x) dx = u sec x dx, so dx = du / (u sec x). Since sec x + tan x = u and sec x - tan x = 1/u, we have 2 sec x = u + 1/u, so sec x = (u^2 + 1)/(2u). The integral becomes \int sec^2 x / u^9/2 dx = \int (sec x) (sec x dx) / u^9/2 = \int ((u^2 + 1)/(2u)) (du / u) / u^9/2 = 1/2 \int (u^2 + 1) / u^(9/2 + 2) du = 1/2 \int (u^2 + 1) / u^13/2 du = 1/2 \int (u^-9/2 + u^-13/2) du = 1/2 [ u^-7/2 / (-7/2) + u^-11/2 / (-11/2) ] + K = - 1/7 u^-7/2 - 1/11 u^-11/2 + K = - 1/u^11/2 [ 1/11 + 1/7 u^2 ] + K = - 1/(sec x + tan x)^11/2 { 1/11 + 1/7 (sec x + tan x)^2 } + K.